用于打开文件和刷新数据连接的VBA语法错误

Question

不确定这里发生了什么......在excel vba中创建一个子程序，打开并通过传递给它的值激活文件。显然我做错了什么......不知道是什么。

Sub openBook(ByVal fName As String, ByVal activate As Boolean)
    Application.Workbooks.Open(fName, 0, False) '= Required here?
End Sub

修改 有了这个工作，只是想检查以确保这是正确的语法如下 请参阅更新的代码：

Sub openBook(ByVal fileName As String, ByVal refresh As Boolean)
    Dim wb As Workbook

    Set wb = Workbooks.Open(fileName, 0, False)

    If refresh = True Then
        wb.RefreshAll
    End If

End Sub

Answer 1

Workbooks.Open的语法是

expression.Open(FileName, UpdateLinks, ReadOnly, Format, Password, WriteResPassword, IgnoreReadOnlyRecommended, Origin, Delimiter, Editable, Notify, Converter, AddToMru, Local, CorruptLoad)

这是你在尝试的吗？

Sub Sample()
    openBook "C:\MyFile.xlsx", False, True
End Sub

Sub openBook(fileName As String, UpdtLink As Boolean, RdOnly As Boolean)
    Application.Workbooks.Open fileName, UpdtLink, RdOnly
End Sub

修改

如果您想通过0/False和1/True，则必须更改

Sub openBook(fileName As String, UpdtLink As Boolean, RdOnly As Boolean)

到

Sub openBook(fileName As String, UpdtLink As Variant, RdOnly As Variant)

从评论中跟进

  

无论如何还要在同一行中激活该工作簿，还是需要另一行代码？ - metsales 1分钟前

为什么要激活它？应尽可能避免.Activate。您可能希望看到THIS

话虽如此，如果你想激活它，那么你必须使用像这样的代码

Sub Sample()
    openBook "C:\MyFile.xlsx", False, True
End Sub

Sub openBook(fileName As String, UpdtLink As Boolean, RdOnly As Boolean)
    Dim wb As Workbook
    Set wb = Application.Workbooks.Open(fileName, UpdtLink, RdOnly)
    wb.Activate
End Sub

但是根据我之前关于不使用.Activate

的建议，以下是我的建议

Dim wb As Workbook

Sub Sample()
    openBook "C:\MyFile.xlsx", False, True

    DoEvents

    With wb
        '
        '~~> Do something with the workbook here
        '
    End With
End Sub

Sub openBook(fileName As String, UpdtLink As Boolean, RdOnly As Boolean)
    Set wb = Application.Workbooks.Open(fileName, UpdtLink, RdOnly)
End Sub

Answer 2

尝试这样的事情（未经测试）

Sub openBook(ByVal fileName As String, ByVal activate As Boolean)
    Application.Workbooks.Open fileName:=filename
End Sub